Find the next number in this sequence:
2, 4, 8, 13, 20, 29, 41, ?
EXTENSION: Find the number for the nth term in the sequence (in terms of n).
Hint:
Fibonacci+Triangle
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5
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$\begingroup$ mathworld.wolfram.com/TribonacciNumber.html $\endgroup$– DuckSep 9, 2018 at 2:33
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$\begingroup$ The hint has nothing to do with that. $\endgroup$– ThyrsoidMouse73Sep 9, 2018 at 2:34
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$\begingroup$ Ok, just checking $\endgroup$– DuckSep 9, 2018 at 2:35
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$\begingroup$ And I changed it to make it more obvious. $\endgroup$– ThyrsoidMouse73Sep 9, 2018 at 2:37
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1$\begingroup$ OEIS found this but I don't think that is what you had in mind $\endgroup$– Bennett BernardoniSep 9, 2018 at 2:45
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2 Answers
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I think the answer is
57
Solution for the extension: (Thanks @PerpetualJ for MathJax)
$$Fi. Δ.$$ $$1 + 1 = 2$$ $$1 + 3 = 4$$ $$2 + 6 = 8$$ $$3 + 10 = 13$$ $$5 + 15 = 20$$ $$8 + 21 = 29$$ $$13 + 28 = 41$$ $$21 + 36 = 57 \leftarrow (Next Term)$$
Round off to the nearest integer.
$Fi(n) = round\biggl(\frac{(-\frac{1}{\phi})^n}{\sqrt{5}}\biggr)$
$\Delta(n) = \frac{n(n + 1)}{2}$
$\int(n) = Fi(n) + \Delta(n) = round\biggl(\frac{(-\frac{1}{\phi})^n}{\sqrt{5}}\biggr) + \frac{n(n + 1)}{2}$
answered Sep 9, 2018 at 5:02
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The next number is
$34$
The formula is
$2^n - \dfrac{(n-1)(n-2)(n-3)}2$
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$\begingroup$ I'm thinking of a different formula and I realized this worked as well so I'll edit the question. $\endgroup$ Sep 9, 2018 at 2:25